Let $G$ be a finite group and $H$ be a nilpotent subgroup of $Aut(G)$. If $C_{G}(H)=1$, is $G$ solvable?
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4$\begingroup$ The question is not well formulated. What do you know about the problem? Works of Shult, Gross, Khukhro and many others? What if $G$ is an Abelian group? $\endgroup$– user6976Apr 30, 2013 at 11:52
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1$\begingroup$ I've been musing on a this question under the (very strong) extra assumption that $G$ is simple, but even this seems a little tricky. I guess that $C_G(H)$ is never trivial here, but how to prove it? One could do an exhaustive analysis of the outer automorphism group, but that seems very unsatisfactory... I may well be missing an easy solution though... $\endgroup$– Nick GillApr 30, 2013 at 15:05
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$\begingroup$ @Nick: The outer automorphism groups of all finite simple groups are well known (and very small). $\endgroup$– user6976Apr 30, 2013 at 16:59
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$\begingroup$ Yes.The outer automorphism groups of all finite simple groups are well known and very small. But how to prove it rigorously, especially when $G$ is a simple group of Lie type? $\endgroup$– sifeApr 30, 2013 at 18:21
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1$\begingroup$ Yes. When $H$ is a p-group, we can use Theorem 6.2.3 in the book "finite groups (gorenstein)" to deduce that $(|G|,|H|)=1$. So by the Classification Theorem of Finite Simple Groups, $G$ is solvable. But I don't know how to check when $G$ is a simple group of Lie type. $\endgroup$– sifeApr 30, 2013 at 21:36
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